It is known that the opening of the quadratic function is upward and the vertex coordinate is (-1, 0).

That is, the product of two roots is 1/a= 1, so a= 1, -b/a=-2 b=2.

f(x)=x^2+2x+ 1

f(x)=x^2+2x+ 1 x & gt； 0

f(x)=-(x^2+2x+ 1)x & lt； 0

(2) when x belongs to increasing function, g (x) = f (x)-kx = x 2+(2-k) x+1,the symmetry axis must be on the left side of the interval, that is.

(k-2)/2 = & lt； -2k & lt； =-2

(k-2)/2 > if the subtraction function needs the symmetry axis on the right side of the interval; = 2k & gt； =6

To sum up, k & lt=-2 or k & gt=6.

(3) If f (x) is an even function, there must be b=0.

f(x)=ax^2+ 1

According to the condition Mn

Suppose m is a positive number and n is a negative number.

Because f(x) is an even function, we can know that f(-x)=f(x).

F(n)=-f(n)=-f(-n)

Because of a>0, the symmetry axis of the function is x=0.

F(m)+F(n)=f(m)-f(-n)

Because m+n >; 0 so m & gt-n & gt；; 0

And f(m) is increasing function in the interval greater than 0, so f (m)-f (-n) >; 0

That is f (m)+f (n) >: 0.